A weak acid,HA,is found to be 10% ionized in | vedclass

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(C) For the weak acid $HA$,the degree of ionization $\alpha = 0.1$ at concentration $C = 0.01 \ M$.The dissociation constant $K_a$ is given by $K_a =

Vedclass · Accepted Answer

$3.653$

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(C) For the weak acid $HA$,the degree of ionization $\alpha = 0.1$ at concentration $C = 0.01 \ M$.The dissociation constant $K_a$ is given by $K_a = \frac{C \alpha^2}{1 - \alpha} = \frac{0.01 	imes (0.1)^2}{1 - 0.1} = \frac{0.0001}{0.9} = 1.11 	imes 10^{-4}$.Calculate $pK_a$: $pK_a = -\log(1.11 	imes 10^{-4}) = 4 - \log(1.11) \approx 4 - 0.0458 = 3.9542$.Using the Henderson-Hasselbalch equation for the buffer solution containing $HA$ $(0.1 \ M)$ and $NaA$ $(0.05 \ M)$:$pH = pK_a + \log \left( \frac{[Salt]}{[Acid]} ight) = 3.9542 + \log \left( \frac{0.05}{0.1} ight)$.$pH = 3.9542 + \log(0.5) = 3.9542 - 0.3010 = 3.6532 \approx 3.653$.

$A$ weak acid,$HA$,is found to be $10\%$ ionized in $0.01\ M$ aqueous solution. Calculate the $pH$ of a solution which is $0.1\ M$ in $HA$ and $0.05\ M$ in $NaA$.

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